A rocket is fired straight up. It contains two stages (Stage 1 and Stage 2) of s

Question

A rocket is fired straight up. It contains two stages (Stage 1 and Stage 2) of solid rocket fuel that are designed to burn for 10.0 and 5.00 s, respectively, with no time interval between them. In Stage 1, the rocket fuel provides an upward acceleration of 18.0 m/s2. In Stage 2, the acceleration is 13.0 m/s2. Neglecting air resistance, calculate the maximum altitude above the surface of Earth of the payload and the time required for it to return to the surface. Assume the acceleration due to gravity is constant.

1) Calculate the maximum altitude.

2) Calculate time required to return to the surface (i.e. the total time of flight).

Explanation / Answer

From the given question,

time for stage 1 (t1)=10s

time for stage 2(t2)=5s

acceleration in stage 1(a1)=18 m/s2

acceleration in stage 2(a2)=13 m/s2

initial velocity in stage 1(u1)=0 m/s

distance travelled in stage 1=s1

s1= u1t1 + (1/2)a1t12

=(0)(10) +(1/2)(18)(10)2

=900m

final velocity after first stage=v1

v1=u1 + a1t1

=0 + 18 (10)

=180 m/s

v1=u2

distance travelled in second stage

s2= u2t2+ (1/2)(a2)(t22)

=(180)(5) + (1/2)(13)(52)

=900 + 162.5

=1062.5m

final velocity after second stage=v2

v2= u2 + a2t2

=180 + (13)(5)

=245 m/s

due to inertia rocket will go upwards of its own against gravity

u3=v2

v3=0

v3=u3 - g(t3)

0=245-(10)t3

t3=24.5

s3=u3t3-(1/2)(g)(t32)

=(245x24.5)-(1/2)(10)(24.52)

=3001.25

maximum altitude=900+1062.5+3001.25=4963.75m

maximum altitude is 4963.75m

then rocket falls under gravity

u4=0, s4=4963.75, a=10

s4=u4t4 + (1/2)(10)(t4)2

4963.75=0 + 5 t42

t4=31.5 sec

total time taken =10+5+24.5+31.5=71

Time required to return to surface is 71 sec.

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